$g_{i}$ and $g_{j}$.
The first step of this stage consists in building the following non-oriented
-graph furthere denoted as to \emph{similarity graphe}.
+graph further denoted as to \emph{similarity graph}.
In this one, the vertices are the genes. There is an edge between
$g_{i}$ and $g_{j}$ if the rate $r_{ij}$ is greater than a given similarity
-treeshold $t$.
+threshold $t$.
We then define the relation $\sim$ such that
$ x \sim y$ if $x$ and $y$ belong in the same connected component.
are also elements of the same equivalence class.
Let us then consider the set of all equivalence classes of the set of genes
by $\sim$, denoted $X/\sim = \{\dot{x} | x \textrm{ is a gene}\}$.
-defined by \pi(x) = \dot{x}
-which maps each gene into it respective equivalence classe by $\sim$.
+defined by $\pi(x) = \dot{x}$
+which maps each gene into it respective equivalence class by $\sim$.
-For each genome $[g_l,\ldot,g{l+m}]$, the second step computes
+For each genome $[g_l,\ldots,g{l+m}]$, the second step computes
the projection of each gene according to $\pi$.
The resulting genome which is
$$
-[\pi(g_l),\ldot,\pi(g{l+m})]
+[\pi(g_l),\ldots,\pi(g{l+m})]
$$
is again of size $m$.
-Intuitivelly speaking, for two genes $g_i$ and $g_j$
+Intuitively speaking, for two genes $g_i$ and $g_j$
in the same equivalence class, there is path from $g_i$ and $g_j$.
It signifies that each evolution step
(represented by an edge in the similarity graph)
Each genome is projected according to $\pi$. We then consider the
intersection of all the projected genomes which are considered as sets of genes
and not as sequences of genes.
-This results as the set of all the class representents $\dot{x}$
-such that each geneome has an gene $x$ in $\dot{x}$.
+This results as the set of all the class $\dot{x}$
+such that each genome has an gene $x$ in $\dot{x}$.
The pan genome is computed similarly: the union of all the
projected genomes in computed here.
